Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
890 N
Explanation:
Acceleration is change in velocity over change in time.
a = Δv / Δt
a = (11 m/s − 0 m/s) / 0.26 s
a = 42.3 m/s²
Force is mass times acceleration.
F = ma
F = (21 kg) (42.3 m/s²)
F ≈ 890 N
Where are the factors ... to this question
As per Bernuolli's Theorem total energy per unit mass is given as

now from above equation




now by above equation


Part B)
Now energy per unit weight


